Calculus- Instantaneous Rate of Change?

Question

The problem is about a door with an automatic closer. At time t=0 seconds someone pushes the door open. It swings open, slows down, stops, then starts closing, it closes slowly until it hits a certain point, when at t=7, it slams shut. As the door is in motion the number of degrees (d) it is from its closed position depends on the time(t).

1. i need help sketching a reasonable graph of d as a function of t.

2. then, suppose that d is a function of t is defined as d(t) = 200t(2^(-t). and sketch that graph

3.then the values of d for each second from t=0 through t=10. (round answers to nearest 0.001)

4. At time t=1 second, does the door appear to be opening or closing? explain. then t=3?

5. is there a time where the door is neither opening or closing? explain

i could really use the help

i am completely lost!

Answer 1

1. as t increases d increases to a maximum of 90 degrees, then as t increases d decreases slowly until t=7, then d drops from where it is to 0.

2. I don't know how to insert a sketch

3. Without more info you can't, you would need to know at what point it slows down, by how much, for how long, how long the door stays open, what rate the door closes at, for how long, and how fast "slams shut" is. Though assuming a perfect world if t=7 means the door is closed, you could say from t=7 to 10, d=0.

4. At t=1 you would not be able to tell if the door is opening or closing unless you knew the doors velocity, you would already know its position (d).

5. the door is neither opening nor closing before t=0, at the point where the door stops in its swing, and after t=7.